BALTIC-WAY 2024 Problem 12

Let $A B C$ be an acute triangle with circumcircle $\omega$ such that $A B<A C$. Let $M$ be the midpoint of the arc $B C$ of $\omega$ containing the point $A$, and let $X \neq M$ be the other point on $\omega$ such that $A X=A M$. Points $E$ and $F$ are chosen on sides $A C$ and $A B$ of the triangle $A B C$, respectively, such that $E X=E C$ and $F X=F B$. Prove that $A E=A F$.